Calculating A Cable Size
For typical domestic lighting and general purpose socket circuits one can usually use one of "standard" circuits presented in the On Site Guide, safe in the knowledge that so long as you follow the guidelines, the combinations of protective device and cable specified will meet all of the (sometimes complex and confusing) requirements applicable. This article is for all those cases where the "off the shelf" circuits are not appropriate, and you are faced with the need to design your own circuit, and prove its adequacy.
Choosing the right sized cable is not always as easy as it looks.
Selecting the correct cable for the application is imperative to ensure a satisfactory life of conductors and insulation subjected to the thermal effects of carrying current for prolonged periods of time in normal service.
Choosing the minimum size cross sectional area of the conductors is essential to meet the requirements for
- Protection against electric shock
- Protection against thermal effects
- Overcurrent protection
- Voltage drop
- Limiting temperatures for terminals of equipment to which the conductors are connected
How to size a cable
Differences between fault current protection and overload current protection
All circuits are designed to protect the cables against fault current (e.g. a short circuit from line to earth or line to neutral). However full overload current protection is not always required (or possible) on some circuits.
Overload protection is required when a cable may conceivably be required to carry a circuit current in excess of its design current for a period of time, either brought about by the circuit's user, or by a predictable failure mode of equipment supplied by the circuit. e.g. a user connecting too many high power appliances to a socket circuit.
Overload current protection is not required when we know the fixed current of the appliance, and we install a dedicated cable that is capable of carrying this current. A good example of this would be an electric shower or an immersion heater supply cable where there is no end user action, or typical equipment failure mode that could result in an overload. We will show in the worked examples later on, that it is even possible in some circumstances to safely install an MCB that has a higher rating than the cable's maximum current rating.
Section 433.3.1 of BS7671 also describes other circumstances where overload current protection can be omitted. Note also there are also some classes of equipment where overload current protection may be omitted on safety grounds (typically fire detection and extinguisher systems, life support equipment, some rotating or lifting equipment etc. See section 433.3.3 of BS7671 for full details).
Overload current protection is required for any circuit where a user could potentially raise the current demand from the circuit to above that anticipated in its design. This would be the case with the vast majority of general purpose socket circuits for example.
Sizing conductors for your circuit
How to calculate Iz
Iz and some other terms we will require shortly are defined as:
- Iz - The rated current carrying capacity of the chosen cable, for continuous service, under the particular installation conditions.
- It - The nominal current carrying capacity of the cable. For T&E cable It is taken from this table.
- Ib - The design current of the circuit. This is the starting point for all the calculations. If available, then Ib should be taken from the rated full load current of the appliance. If this is not given, then Ib may be calculated by dividing the VA rating of the appliance by 230 (the nominal voltage). For cases where the VA rating is not provided, then use the quoted power in Watts divided by 230. Note that diversity can be applied to reduce the full load current where permitted (e.g. with a cooker circuit in a domestic environment)
- In - The rated current of the protective device.
- I2 - For completeness we will include I2 - the actual operating current of the protective device. However for domestic work there are only two classes of protective device of interest: Cartridge Fuses, MCBs, and RCBOs where I2 is 1.45 x In, and rewireable fuses where I2 is 2.0 x In, so we can simplify things a little summaries of the requirements given in case 1 and case 2 below:
Having calculated the easy bits: the design current (Ib) and then protective device rating (In) we now need to calculate Iz.
Iz may be found either by reference to the tables in BS7671 (the wiring regs) or the IEE On Site Guide (a subset of which are reproduced here for common domestic cables sizes), or, by calculation based on the various factors that affect the installation.
Case 1: overload protection required
When overload protection is required, we must ensure that:
- In >= Ib (Overload Protective Device rating is equal to or bigger than load)
and also, for cartridge fuse, MCB or RCBO protective devices that:
- Iz >= In (cable rating (as-installed) is equal to bigger than the Overload Protective Device rating)
or for rewireable fuse then:
- Iz >= In / 0.725 (cable must be up-rated to allow for coarser protection from this type of fuse);
Case 2: overload protection not required, due to characteristics of load
Where overload protection is not required then Iz should ideally be greater than or equal to the MCBs nominal rating, (i.e. Iz >= In). However if this can't easily be achieved, then it is also acceptable to opt for Iz > Ib even if Iz is actually less than the nominal rating of the MCB (i.e. Ib <= Iz <= In). Warning: If the latter design option is used, then it should be remembered that the cable size will have been verified as adequate only for the selected appliance, and it may not be adequate for a more powerful appliance even if the MCB could in theory support it.
In summary: when overload protection is not required, then we must ensure that:
- In >= Ib (as for case 1)
- Iz >= Ib (cable rating (as-installed) equal to bigger than load)
but if Iz < In an adiabatic calculation must be performed to verify fault (i.e. short circuit) protection of the cable is still provided (see section below).
Method 1 - by reference to the tables With direct reference to the table. If the cable is installed ungrouped, in an ambient temperature of 30C, and is protected by a B type MCB, then Iz = It and the cable size is chosen by looking down your reference method column to find Iz. (note ratings for a number of more exotic installation methods can be found in BS7671)
Method 2 - calculation Iz is calculated by using the formula Iz = It x Ca x Cg x Ci x Cc Where It is column C of the table. Ca is a correction factor due to the ambient temperature (values from table 4B1) Cg is a correction value for cables grouped with other circuits (values from table 4C1) Ci is a correction value for cables in insulation (Table 52.2) Cc is a correction factor of 0.725 for BS3036 fuses and 0.9 for cables "in a duct in the ground" or "buried direct". If a buried cable is protected by a BS 3036 fuse then Cc = 0.725 x 0.9 = 0.653
Worked Example Say we have a radial circuit feeding a pair of 3kW immersion heaters. The cable will be grouped with two other circuits and will pass through an aperture in a fully insulated stud wall, containing 100mm of slab insulation. The ambient temperature of the insulated wall is 40°C. The circuit protection will be a B32 MCB, and the cable is initially specced as 4.0mm² T&E. So we know that Ib = 2 x 3000 / 230 = 26A Initial inspection of column C of the table shows a rating for 4.0mm² cable at 37A. However from the tables below we can see that the ambient temperature of 40°C yields a derating of 0.87 and our total of three circuits grouped together gives a factor of 0.7. Finally the 100mm of insulation introduces a further factor of 0.78. Since the protective device is a MCB there is no factor to apply due to the use of a BS 3036 re-wireable fuse. Iz = 37 x 0.87 x 0.7 x 0.78 x 1 = 17.57A Since overload protection is not required for this circuit, we need to achieve only Iz > Ib as a minimum requirement, however in this case it is clear that we have not achieved this. Even drilling extra access holes for the cable to remove the grouping related factor, will still not meet the target. Hence we will have to increase the cable size to 6.0mm² and drill some extra holes: Reworking with the new It of 47A, and removing the grouping factor we get: Iz = 47 x 0.87 x 1 x 0.78 x 1 = 31.89A This does meet the minimum requirement of Iz > Ib and hence is acceptable. It is however very slightly outside the ideal of Ib <= In <= Iz. (a more practical solution may actually be to wire each heater using its own 2.5mm² T&E cable. Since the reduced load on each of 3kW (13A), will come in with a Iz of just over 18A if one also removes the grouping factor). One could even go further by joining this pair of radials to make a ring!
|Ambient temperature||Derating factor Ca|
|Length in insulation (mm)||Dereating factor Ci|
|Arrangement (cables touching)||Number of circuits||Applicable reference method|
|Bunched in air, on a surface, embedded or enclosed||1.0||0.80||0.70||0.65||0.60||0.57||0.54||0.52||0.50||0.45||0.41||0.38||A to F|
|Single layer on a wall||1.0||0.85||0.79||0.75||0.73||0.72||0.72||0.71||0.70||0.70||0.70||0.70||C|
- The Cg value applies to the number of circuits not the number of cables
- If a cable is to be expected to carry less than 30% of it's grouped rating it may be ignored for the purpose of obtaining the rating factor for the rest of the group
Checking Voltage Drop
For lighting circuits the maximum volts drop is 3% (6.9V) and for all other circuits the maximum voltage drop is 5% (11.5V).
We now need to test that the chosen cable will be big enough to supply the circuit without dropping the maximum allowed voltage drop. The table below shows the voltage drop in mili Volts (mV) per amp, per meter. So to find the total drop simply multiply this value by the design current Ib, and the length of the cable L:
Total voltages drop = (Voltage Drop x Ib x L) This will give the voltage drop in mV. Divide by 1000 to convert to Volts
|Conductor CSA (mm²)||PVC (max 70° C)
Voltage drop mV/A/m
Worked Example Say we have an electric shower with a design current Ib of 41A, and a cable run of 19m, installed using Method C. Initial checks would suggest that 6mm² T&E will be adequate. From the table above we know that 6mm² will drop 7.3 mV/A/m. So: Total Drop = 7.3 x 41 x 19 / 1000 = 5.67V which is acceptable
Checking the Maximum Earth Loop Impedance
|Type B circuit breakers to BS EN 60898|
- The circuit loop impedances given in the table should not be exceeded when the conductors are at their operating temperature. If the conductors are at a different temperature when tested, the reading should be adjusted accordingly.
Having now chosen a cable that is suitable to carry our design current and meet the required voltage drop requirements we now need to check that the cable will allow the MCB , fuse or RCBO to trip the circuit quickly enough in the event of a fault.
A fault may be caused by either a line to neutral fault or a line to earth fault. A prospective fault current (PFC) is a line to earth fault limited by the Earth Loop Impedance (ELI) and a prospective short circuit (PSC) is a line to neutral fault limited by the Line Impedance.
Non time delayed RCD protected circuits automatically protect the circuit against line earth faults however it is advisable when possible to design the circuit so that the MCB or overcurrent characteristics of an RCBO still clears the fault in the case of a faulty RCD.
For TN supplies disconnection times are 0.4 seconds for circuits up to 32A and 5 seconds for other circuits. Fortunately for a B type MCB or RCBO the trip current for both 0.4 and 5 seconds disconnection are the same. Table 41.3 above gives the maximum impedance to meet the disconnection times.
To check the disconnection times, we need to know the ELI or LI at the far end of the supply cable. To calculate this we in turn need to know the Earth Loop Impedance or Line Impedance at the consumer unit, this can be found by measurement if you have the appropriate test gear. Alternatively use the default figures of 0.8 for a TN-S earthing system and 0.35 for a TN-C-S system (for TT systems see notes below). Add to this the cable resistance. You can find this by multiplying the length of the cable by the appropriate value from the table below.
|Wire CSA/CPC (mm²)||L + N
|L + CPC |
|1.0 / 1||43.44||43.44|
|1.5 / 1||29.04||36.24|
|2.5 / 1.5||17.78||23.42|
|4.0 / 1.5||11.06||20.05|
|6.0 / 2.5||7.39||12.59|
|10.0 / 4||4.39||7.73|
|16.0 / 6||2.76||5.08|
- The Wire Cross Section Area (CSA) column also indicates the typical CSA of the CPC wire used in a modern cable.
- The values in this table are for copper cables at 70degC and have been increased by a correction factor of 1.2 from the values of those in Table 9A of the OSG to allow for the increased resistance due to operating temperature
If the calculated impedance is less than in table 41.3 then the disconnection times are met.
Worked Example Say we have an electric water heater with a design current Ib of 41A, and a cable run of 19m, installed using Method C. Initial checks would suggest that 6mm² T&E will be adequate. We have already confirmed the voltage drop is acceptable so now we need to check the disconnection times. From the above table ELI = 19 x 12.59 / 1000 = 0.24 ohms and LI = 19 x 7.39 / 1000 = 0.14 ohms For a TN-S supply with a Ze of 0.8 ohms then the calculated maximum resistances are ELI is 0.8 + 0.24 = 1.04 ohms LI is 0.8 + 0.14 = 0.94 ohms As the ELI of 1.04 ohms is greater than the maximum 1.02 ohms allowed by a 45A MCB then this cable installation is not suitable for use on a TN-S supply without RCD protection. Alternatively specifying a larger CSA cable (or shortening the route of the existing one) may reduce the ELI sufficiently to bring the design in spec without needing to rely on the RCD.
TT Earthing systems will rarely provide a low enough ELI to meet disconnection times (or for that matter allow a sufficiently large PFC to flow to achieve disconnection at all!) Hence they must rely on RCD protection on all circuits to achieve earth fault disconnection. Note also that the disconnection time limits are tighter for TT systems, at 0.2 secs for circuits up to 32A, and 1 second for other circuits. (typical RCDs will disconnect in approx 40ms for "normal" types, and 900ms for Time delayed (Type S) devices).
The final design exercise is to check that in the event of a fault (i.e. very high current short circuits between line and earth or line to neutral etc), the cable has sufficient conductor cross sectional area to survive long enough to allow the circuit breaker or fuse to clear the fault without the cable being damaged by overheating. As a starting point one needs to assume that the cable is already running at its maximum design temperature (e.g. 70°C for PVC T&E). The high short circuit current will result in rapid heating of the cable, and given that this will happen very quickly there will be little time for any of this energy to be dissipated to the cables surroundings. This is known as adiabatic heating. This is compounded further by the fact that the wire resistance will rise with temperature, and hence the heating effect will also increase in direct proportion.
The wiring regs handle this situation with what is known as the adiabatic equation.
The equation for this is usually arranged to calculate the minimum required conductor cross section "s":
- s = sqrt( I² x t ) / k
Where I is the prospective fault current, and t is the time to open the circuit breaker (typically 0.1 secs) k is a constant that takes into account the characteristics of the materials it is made from as well as highest possible short term rise in conductor temperature that it will tolerate without damage. See table below for a list of common values (or see BS7671 table 43.1 for other cables not covered here):
k values for copper conductor cables of CSA < 300mm²
|Copper conductors with
|70°C Thermoplastic (general purpose PVC)||70||160||115|
|90°C Thermoplastic (PVC)||90||160||100|
|60°C Thermosetting (eg Rubber)||60||200||141|
|90°C Thermosetting (eg XLPE)||90||250||143|
Worked Example If we take the previous example of a circuit with 19m of 10mm² T&E on a TN-S supply with a ELI of 0.8 ohms we can calculate the ELI at the far end as 0.8 + (19 x 7.73 / 1000) = 0.95 ohms. So we can now calculate the prospective fault current as 230 / 0.95 = 243A, which will open a 45A MCB in 0.1 secs. So using these figures: s = sqrt( 243² x 0.1 ) / 115 = 0.67mm² Since the CPC in 10mm² T&E is 4mm² this will be more than adequate.